 Solution

At $\alpha =1$, the phase is $-x+\sin x=-x^3/6+O(x^5)$, so the ordinary nonstationary endpoint formula fails. In the degenerate endpoint in Laplace's method, take $x={\lambda }^{-1/3}t$ and extend the localized integral to infinity:The last equality follows by $u=t^3/6$. It is the $\nu =0$ value of the cubic Laplace transition integral after its normalization is rescaled to the coefficient $1/6$.